and we know \[\frac{d}{dx} \frac{e^{-x}}{1-e^{-x}} \\ \frac{d}{dx} \frac{1}{e^x-1} \\ \frac{d}{dx}(e^x-1)^{-1} \\ -1(e^x-1)^{-2}(e^x) \\ - \frac{e^x}{(e^x-1)^2}\]
but we should evaluate this at x=1 since we had that we really wanted to evaluate
\[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]
\[(e-1) \frac{-e}{(e-1)^2}=\frac{-e}{e-1}\]
but I think we are to get 1/(e-1)

thanks @ganeshie8
ok I was sorta failing at the algebra a little too
but I see it now

freckles

2 years ago

I can't remember who gave me this problem but @Loser66 this is also on your gre practice exam if you want to look at this.

ganeshie8

2 years ago

i remember working this problem differently sometime back here in openstudy but google isn't helping at the moment..

freckles

2 years ago

And I bet you have answered so many questions since this that going back through your old questions is horrifying and it also doesn't work to well sometimes. Sometimes it automatically throws me out of the looking through my old question thing. Not sure if you know what I mean.

Loser66

2 years ago

Thanks so mmmmmmmmuch

xapproachesinfinity

2 years ago

truly i could not see how you threw differentiation in there @gabylovesu
how is that sum equal to that differentiation ?

xapproachesinfinity

2 years ago

@ganeshie8

xapproachesinfinity

2 years ago

wrong tagging lol

anonymous

2 years ago

Have you tried parameterizing and applying the Laplace transform? I'm thinking something along the lines of
\[I(s)=\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx=\mathcal{L}\{\lfloor x\rfloor\}\]
The transform definitely exists because \(\lfloor x\rfloor\) is piecewise continuous and of exponential order.

ganeshie8

2 years ago

\[\color{red}{\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=-\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}}=-\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= -\frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\]
I think it would be easier to make sense of it by differentiating \(\large \color{red}{e^{-nx}}\) and seeing that you get back the starting expression...

freckles

2 years ago

hey @SithsAndGiggles I haven't thought of that but I would be interested in seeing that way if you wanted to show that way. I honestly I haven't seen any laplace transform action in like 10 years so I kind of forgot all of that. :p

anonymous

2 years ago

The idea would be to express \(\lfloor x\rfloor e^{-x}\) in terms of the step function
\[\theta(x-c)=\begin{cases}1&\text{for }x\ge c\\0&\text{for }x

does seem similar but still pretty @SithsAndGiggles
anyways this is making me hungry
so peace and thanks for the fun

freckles

2 years ago

I will come back later though to analyze more deeply what you said

anonymous

2 years ago

As for the actual result:
\[\sum_{c=1}^\infty c(e^{-c}-e^{-(c+1)})=\sum_{c=1}^\infty c\left(\frac{1}{e^c}-\frac{1}{e^{c+1}}\right)=\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots=\frac{1}{1-\frac{1}{e}}-1\]